What is the integral from 2 to 5 of xe^{2x} ?

The integral from 2 to 5 of xe^{2x} is (9e^{10}-3e^4)/4

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integral from 2 to 5 of xe^{2x}

\int_{2}^{5} x e^{2 x} d x

\frac{9 e ^{10} - 3 e ^{4}}{4}

Solution steps

\int_{2}^{5} x e^{2 x} d x

Apply u-substitution

: \int_{4}^{10} \frac{e ^{u} u}{4} d u

= \int_{4}^{10} \frac{e ^{u} u}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{4}^{10} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{4}^{10}

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{4}^{10}

\int e^{u} d u = e^{u}

= \frac{1}{4} [e^{u} u - e^{u}]_{4}^{10}

Compute the boundaries:

9 e^{10} - 3 e^{4}

= \frac{1}{4} (9 e^{10} - 3 e^{4})

Simplify

= \frac{9 e ^{10} - 3 e ^{4}}{4}

\int_{5}^{2} 1

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\int_{0}^{\frac{9}{8}} \frac{1}{64 x ^{2} + 81} d x

\int_{0}^{1} \frac{2}{( x - 1 ) ^{2}} d x

\int_{0}^{1} 16 x^{2} + 1 d x

What is the integral from 2 to 5 of xe^{2x} ?
The integral from 2 to 5 of xe^{2x} is (9e^{10}-3e^4)/4